Building periodic sequences from finite-support sequences

Question

Given a discrete-time finite-support signal x[n]

$$x[n] = \left\{ {\begin{array}{*{20}{l}} {{{( - 1)}^n}n}&{{\rm{ }}n = 1,2,3}\\ 0&{{\rm{otherwise}}} \end{array}} \right.$$

And consider also its periodic repetition is

$$y[n] = \sum\nolimits_{k = - \infty }^\infty {x[n + 7k]} $$

How we can generate the y[n] in Matlab or Python?

Here is piece of code to generate and plot x[n]. But for Y[n] I have no idea!

% finite support signal
n = -5:1:8;
x = zeros(size(n));
x( find(n==1 | n==2 | n==3) ) = [(-1)^1*1 (-1)^2*2 (-1)^3*3];
stem(n, x, 'filled')

Answer 1

You're thinking too complicated. Sorry I wasn't clear enough in my comment.

You're given a sequence $x[n]$ that contains three spikes and zeros elsewhere. Your sequence $y[n]$ contains periodically shifted copies of the original one. It's just repeating the same three spikes indefinitely, without overlap since $7>3$. Hence, all you need to do to find out what $y[n]$ is for a given $n$ is to check where it would map to in your original sequence. In other words: $$y[n] = x[n \; {\rm mod} \; 7].$$

Here is how you can see why that is true: take $y[n] = \sum_k x[n+7k]$, expand, and then use the fact that $x[n]$ is zero outside $[0,7)$ (in fact, zero outside $[1,2,3]$):

$$\begin{align}y[0] & = x[0] + x[7] + x[14] + \ldots + x[-7] + x[-14] + \ldots = x[0] \\ y[1] & = x[1] + x[8] + x[15] + \ldots + x[-6] + x[-13] + \ldots = x[1]\\ & \ldots\\ y[6] & = x[6] + x[13] + x[20] + \ldots + x[-1] + x[-8] + \ldots = x[6] \\ y[7] & = x[7] + x[14] + x[15] + \ldots + x[0] + x[-7] + \ldots = x[0] \\ y[8] & = \ldots = x[1] \\ & \ldots \\ y[13] & = \ldots = x[6] \\ y[14] & = \ldots = x[0] \\ \end{align}$$

And that's how I would do it in Matlab

x = @(n) (-1).^n .* n .* ((n>=1)&(n<=3));
y = @(n) x(mod(n,7));
n = -7:21;
figure(1);clf;stem(n,x(n))
hold on;
stem(n,y(n),'rx--')
xlabel('n');axis([-7,21,-4,4])
legend('x[n]','y[n]');

*edit: Regarding your question 1: yes, you can do it with a sum as well. If you do it correctly, the result is exactly the same. Here's a take in Matlab: Of course you cannot do an infinite sum, so this example sums from -K to K, plotting for K=10:

x = @(n) (-1).^n .* n .* ((n>=1)&(n<=3));
y = @(n) x(mod(n,7));
y_sum = @(n,K) sum(x(n + 7*(-K:K)'),1);
n = -7:21;
figure(1);clf;stem(n,x(n))
hold on;
stem(n,y(n),'rx--')
stem(n,y_sum(n,10),'k*-.')
xlabel('n');axis([-7,21,-4,4])
legend('x[n]','y[n]');

Regarding your second question, well, the code is there, you can experiment with what changes. If the pulses overlap, you're going to see the sums of the overlapping ones... in particular the -1 and the -3 pulse add up to -4 pulses. Looks like this (now the modulo method does not work like that, since it assumed no overlap):

Answer 2

Thanks to @Florian's comments, here is sample solution

clc, clear
n = 0:1:7*10^6-1;
x = zeros(size(n));
x( find(n==1 | n==2 | n==3) ) = [(-1)^1*1 (-1)^2*2 (-1)^3*3];
y = x(mod(n,7)+1);
y = sum(reshape(y,7,[]),2)'

For large $k$s we have for example: y = 0 -1000000 2000000 -3000000 0 0 0.